Base | Representation |
---|---|
bin | 10010111011001001001… |
… | …011011011110000111001 |
3 | 11121022200222011011210211 |
4 | 102323021023123300321 |
5 | 132301313031403423 |
6 | 2433230531111121 |
7 | 162645332004445 |
oct | 22731113336071 |
9 | 4538628134724 |
10 | 1300455341113 |
11 | 46157a156765 |
12 | 190053a424a1 |
13 | 9582b544256 |
14 | 46d29c37825 |
15 | 23c63dbb10d |
hex | 12ec92dbc39 |
1300455341113 has 2 divisors, whose sum is σ = 1300455341114. Its totient is φ = 1300455341112.
The previous prime is 1300455341089. The next prime is 1300455341189. The reversal of 1300455341113 is 3111435540031.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 666922122409 + 633533218704 = 816653^2 + 795948^2 .
It is a cyclic number.
It is not a de Polignac number, because 1300455341113 - 225 = 1300421786681 is a prime.
It is not a weakly prime, because it can be changed into another prime (1300455341413) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 650227670556 + 650227670557.
It is an arithmetic number, because the mean of its divisors is an integer number (650227670557).
Almost surely, 21300455341113 is an apocalyptic number.
It is an amenable number.
1300455341113 is a deficient number, since it is larger than the sum of its proper divisors (1).
1300455341113 is an equidigital number, since it uses as much as digits as its factorization.
1300455341113 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 10800, while the sum is 31.
The spelling of 1300455341113 in words is "one trillion, three hundred billion, four hundred fifty-five million, three hundred forty-one thousand, one hundred thirteen".
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